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Question

If $x=\mathrm{sin}\left(\frac{1}{a}\mathrm{log}y\right)$, show that (1 − x2)y2 − xy1 − a2y = 0.

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Solution

Here, $x=\mathrm{sin}\left(\frac{1}{a}\mathrm{log}y\right)\phantom{\rule{0ex}{0ex}}⇒\frac{1}{a}\mathrm{log}y={\mathrm{sin}}^{-1}x\phantom{\rule{0ex}{0ex}}⇒y={e}^{a{\mathrm{sin}}^{-1}x}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.x,\mathit{}\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}{y}_{1}={e}^{a{\mathrm{sin}}^{-1}x}×\frac{a}{\sqrt{1-{x}^{2}}}\phantom{\rule{0ex}{0ex}}⇒{y}_{1}=\frac{ay}{\sqrt{1-{x}^{2}}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Differentiating}\mathrm{again}\mathrm{w}.\mathrm{r}.\mathrm{t}.x,\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}{y}_{2}=\frac{a{y}_{1}\sqrt{1-{x}^{2}}+\frac{xay}{\sqrt{1-{x}^{2}}}}{\left(1-{x}^{2}\right)}\phantom{\rule{0ex}{0ex}}⇒{y}_{2}=\frac{a{y}_{1}\left(1-{x}^{2}\right)+xay}{\left(1-{x}^{2}\right)\sqrt{1-{x}^{2}}}\phantom{\rule{0ex}{0ex}}⇒{y}_{2}=\frac{a{y}_{1}}{\sqrt{1-{x}^{2}}}+\frac{xay}{\left(1-{x}^{2}\right)\sqrt{1-{x}^{2}}}\phantom{\rule{0ex}{0ex}}⇒{y}_{2}=\frac{{a}^{2}y}{1-{x}^{2}}+\frac{x{y}_{1}}{\left(1-{x}^{2}\right)}\phantom{\rule{0ex}{0ex}}⇒\left(1-{x}^{2}\right){y}_{2}-x{y}_{1}-{a}^{2}y=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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